I have to prove that $\ell^{\infty}(\mathbb N)$ is not separable.
My attempt
Consider a SUBSET $V$ of $\ell^{\infty}(\mathbb N)$ consisting of bounded sequences that have only $0$, $1$ entries, e.g. $(0,1,1,0,0,0,0,1,0,0,\dots)$
Now assume that this SUBSET is uncountable.
If we take a radius $r=1/4$ then balls with origins that are elements of $V$ would be disjoint and because any base of $\ell^{\infty}(\mathbb N)$ must contain a subset of each element of the set of these balls, base of $\ell^{\infty}(\mathbb N)$ can't be countable so it doesnt satisfy the second axiom fo countability and thus is not separable (since $\ell^{\infty}(\mathbb N)$ is a metric space).
Could someone check this? And I still need to prove that $V$ is uncountable…
Best Answer
As the OP guessed, the right subset to consider is the set $A\subset\ell^\infty$ consisting of all the sequences with zeros and ones, i.e., $\{a_n\}\in A$ if and only if $a_n\in\{0,1\}$, for all $n$.
Clearly, if $\{a_n\},\{b_n\}\in A$ and $\{a_n\}\ne\{b_n\}$, then $$ \|\{a_n\}-\{b_n\}\|_\infty=1, $$ since there exists at least one $n$, for which $a_n\ne b_n$, and hence $|a_n-b_n|=1$.
Also $|A|=\big\lvert 2^{\mathbb N}\big\rvert= 2^{\aleph_0}$. Hence $A$ is uncountable, and in fact equinumerous to $\mathbb R$.
The open sets $$ \big\{B\big(\{a_n\},\tfrac{1}{3}\big): \{a_n\}\in A\big\}, $$ are uncountably many mutually disjoint open sets. The fact that such a collection of open sets exists means that $\ell^\infty$ is not separable.